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polet [3.4K]
3 years ago
12

Your front teeth can act as a simple machine. Which machine are they?

Physics
2 answers:
lawyer [7]3 years ago
4 0
<span>Wedges is your answer please mark brainliest </span>
Leni [432]3 years ago
4 0
Your front teeth act as a wedge for a machine
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Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

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mg ( sin39 - μ cos39 ) = 0

μ = tan39

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net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

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A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi
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Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

\Delta x_1=0.50m-0.35m=0.15m

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

k\Delta x_1=m_1g

Solving for the spring constant k, we get:

k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m

Finally, we sum this to the unstretched length to obtain the length of the spring:

x_2=0.35m+0.088m=0.44m

In words, the length of the spring when the second block is hung from it, is 0.44m.

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